Determine if the following integral diverges/converges, if it converges, is it absolutely or conditionally converges.
$$
\int_{1}^{2}\frac{dx}{\sqrt{x^2-x+1} – 1}
$$
What i tried:
In that interval we know that:
$$
0 < x
$$
Therefore we can write:
$$
\frac{1}{\sqrt{x^2-x+1} – 1} < \frac{1}{\sqrt{x^2-2x+1} – 1} = \frac{1}{\sqrt{(x-1)^2} – 1}
$$
$$
= \frac{1}{x – 2}
$$
Now lets try to calculate the integral, using limits, we get:
$$
\lim_{t \to 2^-}\int_{1}^{t}\frac{1}{x-2} = ln(x-2)|_{1}^{t} = ln(t-2) – ln(1-2)
$$
I cant calculate this integral, namely i dont get any limit that is a number.
I thought that maybe i will get to a form like this:
$$
\int_{a}^{b}\frac{1}{(b-x)^\alpha}, \int_{a}^{b}\frac{1}{(x-a)^\alpha}
$$
But also, if i will, and conclude that $ \int \frac{1}{x – 2}$ diverges in that interval, i couldnt use the comparison test to conclude about the original function.
Can i have a hint?
Thank you.
Best Answer
A quick way to decide about convergence/divergence is the limit comparison test:
$$\lim_{x\to 1^+}\frac{\sqrt{x^2-x+1}-1}{x-1}= \left. \left(\sqrt{x^2-x+1}\right)'\right|_{x=1}$$ $$ =\left. \frac{2x-1}{2\sqrt{x^2-x+1}}\right|_{x=1} = \frac 12$$
Since $\int_1^2\frac{dx}{x-1}$ is divergent, $\int_{1}^{2}\frac{dx}{\sqrt{x^2-x+1} - 1}$ must be divergent, as well.